Problem: Evaluate the definite integral. $\int^{4}_{2}\left(\dfrac{6+x^2}{x^3}\right)\,dx = $ Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{9}{16}$ (Choice B) B $\ln\left(2\right)+\dfrac{9}{16}$ (Choice C) C $-\ln\left(2\right)+\dfrac{3}{16}$ (Choice D) D None of the above
First, simplify and use the power and natural log rules: $\begin{aligned}\int^{4}_{2}\left(\dfrac{6+x^2}{x^3}\right)\,dx ~&=~\int^{4}_{2}\left(\dfrac{6}{x^3}+\dfrac{x^2}{x^3}\right)\,dx \\&=~\int^{4}_{2}\left(6x^{-3}+\dfrac{1}{x}\right)\,dx \\&=\left(-3x^{-2}+\ln(x)\right)\Bigg|^{4}_{2}\end{aligned}$ Second, plug in the limits of integration: $(\ln({4})-3\cdot{4}^{-2})-(\ln({2})-3\cdot{2}^{-2}) = \ln\left(2\right)+\dfrac{9}{16}$. The answer: $\int^{4}_{2}\left(\dfrac{6+x^2}{x^3}\right)\,dx ~=~\ln\left(2\right)+\dfrac{9}{16}$